Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 68 / 5

There are $n$ basis splines for the stated $n+k$ knots; the upper limit $k$ in the printed basis list is an indexing slip. LetThe strictly increasing sites and $t_i<x_i^{\ast }<t_{i+k}$ give invertibility by the Schoenberg–Whitney theorem. The needed additional matrix property is total nonnegativity of B-spline collocation matrices: every ordered minor of $A$ is nonnegative.

Here is a structural justification of that sign property. Knot insertion refines spline coefficients by rules of the form $b_j={\alpha }_j{a}_j+(1-{\alpha }_j)a_{j-1}$, $0\le {\alpha }_j\le 1$, with the usual unchanged coefficients before the insertion and shifted coefficients after it. Each coefficient map is a nonnegative rectangular bidiagonal matrix; all its ordered minors are nonnegative, as its nonzero terms preserve row/column order. Products preserve this property by the Cauchy–Binet formula. Insert each ordered sampling site until it has full knot multiplicity. Evaluation at such a break is one of the refined coefficients, with a consistent one-sided convention. Consequently the B-spline collocation matrix is an ordered row submatrix of the product of refinement matrices, proving its nonnegative minors. This also applies to repeated admissible original knots.

Since $A$ is invertible and totally nonnegative, $\det A>0$. The adjugate identity gives the checkerboard inverse of a totally nonnegative matrix:Now for any spline $s={\sum }_j{a}_j{N}_j$, its sampled-value vector is $y=Aa$. Hence the coefficient linear functional in the dual basis is exactlyIn particular ${\mu }_i(N_{\ell })={\delta }_{i\ell }$, verifying the hinted formula rather than assuming it. The operator norm of this linear functional is taken for the uniform norm on the spline's interval, containing all sampling sites. Evaluation at a sampling site has norm at most one, soFor the given unit-norm alternating spline, $s_{\ast }(x_j^{\ast })=(-1{)}^j$, and the checkerboard signs make all terms in a fixed inverse row align:Since $\Vert s_{\ast }{\Vert }_{\infty }=1$, this supplies the reverse operator norm inequality and proves the Chebyshev spline coefficient and dual norm equalityThis spline attains the operator norm of each coefficient linear functional.

Use the order of the three Bernstein basis functions given in this part. Endpoint evaluation givesso both endpoint coefficients have absolute value at most one. At the midpoint,The triangle inequality therefore gives the quadratic Bernstein coefficient boundAll three bounds are sharp. Indeed $p(x)=8x^2-8x+1=T_2(2x-1)$ has uniform norm one and coefficient vector $(1,-3,1)$ in this basis. The reversed ordering of endpoint Bernstein functions does not affect these symmetric endpoint bounds.